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Homework Log Wed 4/27 Lesson Rev Learning Objective:

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1 Homework Log Wed 4/27 Lesson Rev Learning Objective:
To remember everything on conics Hw: STUDY! Stamp log due Thurs.

2 4/27/16 Chapter 10 Conics Algebra II

3 Learning Objective To graph conics To write equations of conics

4 Write an equation of a parabola with the given information
1. Vertex (1, 4), Focus (1, -2) V(h, k) V c = -6 𝑦= 1 4𝑐 (π‘₯βˆ’β„Ž) 2 +π‘˜ F 𝑦= 1 4(βˆ’6) (π‘₯βˆ’1) 2 +4 𝑦=βˆ’ (π‘₯βˆ’1)

5 Write an equation of a parabola with the given information
2. Focus (2, 3), Directrix y = -1 F 𝑦= 1 4𝑐 (π‘₯βˆ’β„Ž) 2 +π‘˜ c = 2 D 𝑦= 1 4(2) (π‘₯βˆ’2) 2 +1 V(2, 1) 𝑦= π‘₯βˆ’ V(h, k)

6 Write an equation of a parabola with the given information
3. V(0, -3), Directrix x = 1.25 c = -1.25 V(h, k) π‘₯= 1 4𝑐 (π‘¦βˆ’π‘˜) 2 +β„Ž V π‘₯= 1 4(βˆ’1.25) (𝑦+3) 2 +0 D π‘₯=βˆ’ 1 5 (𝑦+3) 2

7 Get into vertex form. Find vertex, focus, and directrix. Sketch a graph.
4. y=2 π‘₯ 2 +8π‘₯+11 V(-2, 3) 𝑦=2 π‘₯ 2 +4π‘₯+_____ +11+____ 4 βˆ’8 D: y = 2 7 8 𝑏 2 2 = = 2 2 =4 𝑦=2 (π‘₯+2) 2 +3 F D V 𝑦= 1 4𝑐 (π‘₯βˆ’β„Ž) 2 +π‘˜ 2 1 = 1 4𝑐 8𝑐=1 𝑐= 1 8 Opens β€œy” up

8 D Get into vertex form if it isn’t already. Find vertex, focus, and directrix. Sketch a graph. 5. π‘₯=βˆ’ 𝑦 π‘₯= 1 4𝑐 (π‘¦βˆ’π‘˜) 2 +β„Ž V(5, -2) βˆ’ 1 8 = 1 4𝑐 F(3, -2) V F 4𝑐=βˆ’8 𝑐=βˆ’2 D: x = 7 Opens β€œx” left

9 Write an equation of a circle with the given information
6. Center (-4, 0), radius 9 (π‘₯βˆ’β„Ž) 2 + (π‘¦βˆ’π‘˜) 2 = π‘Ÿ 2 (π‘₯βˆ’(βˆ’4)) 2 + (π‘¦βˆ’0) 2 = 9 2 (π‘₯+4) 2 + 𝑦 2 =81

10 Write an equation of a circle with the given information
(h, k) 7. Center (-3, 2), point on the circle (1, 5) π‘Ÿ= (1βˆ’(βˆ’3)) 2 + (5βˆ’2) 2 π‘Ÿ= (4) 2 + (3) 2 π‘Ÿ=5 (π‘₯βˆ’β„Ž) 2 + (π‘¦βˆ’π‘˜) 2 = π‘Ÿ 2 (π‘₯βˆ’(βˆ’3)) 2 + (π‘¦βˆ’2) 2 = 5 2 (π‘₯+3) 2 + (π‘¦βˆ’2) 2 =25

11 Write an equation of a circle with the given information
8. Endpoints of a diameter (2, 1) & (-2, 3) π‘Ÿ= 1 2 𝑑 π‘Ÿ= (3βˆ’1) 2 + (βˆ’2βˆ’2) 2 π‘Ÿ= (2) 2 + (βˆ’4) 2 = = = 5 𝐢= π‘₯ 1 + π‘₯ 2 2 , 𝑦 1 + 𝑦 2 2 = βˆ’2+2 2 , 3+1 2 center=(0, 2) (h, k) (π‘₯βˆ’β„Ž) 2 + (π‘¦βˆ’π‘˜) 2 = π‘Ÿ 2 π‘₯βˆ’ π‘¦βˆ’2 2 = π‘₯ 2 +( π‘¦βˆ’2) 2 =5

12 Find the center & radius of the circle & graph
9. π‘₯ 2 + 𝑦 2 βˆ’2π‘₯+8𝑦=9 ( π‘₯ 2 βˆ’2π‘₯+____)+( 𝑦 2 +8𝑦+____)=9+____ +____ 1 16 1 16 𝑏 2 2 = βˆ’ = βˆ’1 2 =1 𝑏 2 2 = = 4 2 = 16 (π‘₯βˆ’1) 2 +( 𝑦+4) 2 =26 Center: (1, -4) Radius: 26 β‰ˆ5.1

13 Find the center, vertices, co-vertices & foci
Find parts Sketch Graph Look at Graph 10. π‘₯ 𝑦 =1 Major Axis: y a = 5 b = 2 Center (0, 0) V(0,Β±5) CV(Β±2,0) F(0,Β± 21 ) 𝑐 2 = π‘Ž 2 βˆ’ 𝑏 2 𝑐 2 = 5 2 βˆ’ 2 2 𝑐 2 =21 c= 21

14 Write an equation of an ellipse with the given information
11. Co-vertices (0, Β±3), major axis length 8 Major Axis: x Sketch Graph Find parts (look at graph) a = 4 b = 3 b = 3 (π‘₯βˆ’β„Ž) 2 π‘Ž (π‘¦βˆ’π‘˜) 2 𝑏 2 =1 Plug in (π‘₯βˆ’0) (π‘¦βˆ’0) =1 a = 4 π‘₯ 𝑦 2 9 =1

15 Find the center, vertices, co-vertices & foci
12. π‘₯ 2 9 βˆ’ 𝑦 =1 Opens: x a = 3 b = 5 Center (0, 0) V(Β±3, 0) F(Β± 34 ,0) Asym: y=Β± 5 3 π‘₯ 𝑐 2 = π‘Ž 2 + 𝑏 2 𝑐 2 = 𝑐 2 =34 c= 34

16 Write an equation of a hyperbola with the given information
13. Foci (0, Β±4), Vertices (0, Β±3) Center is between vertices Center (0, 0) Opens: y c = 4 a = 3 c = 4 𝑐 2 = π‘Ž 2 + 𝑏 2 a = 3 (π‘¦βˆ’π‘˜) 2 π‘Ž 2 βˆ’ (π‘₯βˆ’β„Ž) 2 𝑏 2 =1 4 2 = 𝑏 2 (π‘¦βˆ’0) 2 (3) 2 βˆ’ (π‘₯βˆ’0) =1 b= 7 𝑦 2 9 βˆ’ π‘₯ 2 7 =1

17 Identify the conic section, get into standard form, & graph
14. 4 π‘₯ 2 βˆ’ 𝑦 2 βˆ’16π‘₯βˆ’2π‘¦βˆ’1=0 Hyperbola (4 π‘₯ 2 βˆ’16π‘₯+____)+ βˆ’ 𝑦 2 βˆ’2𝑦+____ =1+____+____ 4 π‘₯ 2 βˆ’4π‘₯+____ βˆ’ 𝑦 2 +2𝑦+____ =1+____+____ 4 1 16 -1 𝑏 2 2 = βˆ’ 𝑏 2 2 = = βˆ’2 2 =4 = 1 2 =1 4 (π‘₯βˆ’2) 2 βˆ’ (𝑦+1) 2 =16 (π‘₯βˆ’2) 2 4 βˆ’ (𝑦+1) =1 Center (2, -1) Opens: x

18 #14 Cont’d (π‘₯βˆ’2) 2 4 βˆ’ (𝑦+1) 2 16 =1 𝑐 2 = π‘Ž 2 + 𝑏 2 𝑐 2 = 2 2 + 4 2
Opens: x a = 2 b =4 (look at graph) Center (2, -1) V(0, -1) (4, -1) F(2Β±2 5 ,βˆ’1) 𝑐 2 = π‘Ž 2 + 𝑏 2 𝑐 2 = c=2 5 𝑐 2 =20 Add along x

19 Identify the conic section, get into standard form, & graph
15. π‘₯ 2 +4 𝑦 2 +6π‘₯βˆ’32𝑦+57=0 Ellipse π‘₯ 2 +6π‘₯+____ + 4𝑦 2 βˆ’32𝑦+____ =βˆ’57+____+____ π‘₯ 2 +6π‘₯+____ +4 𝑦 2 βˆ’8𝑦+____ =βˆ’57+____+____ 9 16 9 64 𝑏 2 2 = 𝑏 2 2 = βˆ’ = 3 2 =9 = βˆ’4 2 =16 (π‘₯+3) 2 +4( π‘¦βˆ’4) 2 =16 (π‘₯+3) ( π‘¦βˆ’4) 2 4 =1 Center (-3, 4) Major Axis: x

20 #15 Cont’d (π‘₯+3) ( π‘¦βˆ’4) 2 4 =1 Major Axis: x a = 4 b = 2 (look at graph) Center (-3, 4) V(1, 4) (-7, 4) CV (-3, 2) (-3, 6) F(-3Β±2 3 , 4) 𝑐 2 = π‘Ž 2 βˆ’ 𝑏 2 𝑐 2 = 4 2 βˆ’ 2 2 c=2 3 𝑐 2 =12 Add along x


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